The ph of 0.010 m aqueous aniline is 8.32 what is the percentage protonated. we will discuss how to determine the value of percentage protonated. and more.

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The ph of 0.010 m aqueous aniline is 8.32. what is the percentage protonated.

first off all we will discuss about pOH,

We can find out the value of pOH with the help of following equation. Such as,

pH + pOH – 14

pOH = 14 – ph

POH = 14 – 8.32

pOH = 5.68

Now, we will discuss about hydroxide ion concentration,

we know that,

pOH = -log [OH-]

You can find hydroxide ion concentration with the help of this equation.

The equation is,

pOH = -log [OH-]

now,

5.68 = -log[OH-]

[OH-] = 2.09 × 10×10×10×10×10×10

Now, you know that

Concentration of OH- ion = concentration of NH4+

= 2.09 × 10×10×10×10×10×10

Now, you have to calculate the percentage aniline protonated

= 2.09 × 10 ×10×10×10×10×10/ 0.010 m

= 0.0209%

the final results is,

Percentage aniline protonated = 0.0209%

Read More:- azo dye test for aromatic amines?

Predict the organic products formed when bzcl reacts with aniline. bzcl = benzoyl chloride.

Organic products is formed, when bzcl is react with aniline, where bzcl = benzoyl chloride,

the reaction is,

Aniline + benzoyl chloride – ?

aniline react with benzoyl chloride

in this reaction, Nitrogen has lone pair, so it attack on electrophile (COCl). and it is given product as shown in the reaction.

After this reaction, N-phenylbenzyamide and HCl is removed

The organic products is, N-phenylbenzyamide.

Organic products

what is the name of this organic products?

Answer – N-phenylbenzyamide


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